Proof 3

Proof 4

For $n=1\,$ there is nothing to prove. Let's check the case of $n=2.\,$ We have $a_1a_2=1\,$ and need to prove that $(1+a_1)(1+a_2)\ge 4.\,$ But

$(1+a_1)(1+a_2)=1+a_1+a_2+a_1a_2=2+(a_1+a_2)\ge 2+2\sqrt{a_1a_2}=4.$

Assume the truth of the statement for some $n\ge 2\,$ and consider $n+1\,$ positive numbers such that $\displaystyle\prod_{k=1}^{n+1}a_k=1.\,$ One consequence of that condition is that there are bound to be two numbers, say, $a_{n}\le 1\,$ and $a_{n+1}\ge 1.\,$ By the inductive assumption, $\displaystyle\prod_{k=1}^{n-1}(1+a_k)(1+a_{n}a_{n+1})\ge 2^{n}.\,$ We thus have